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I will give out all the details in case it is relevant. I have a MIMO state space system. I find its bode plot using MATLAB and separately using Mathematica. The plot from MATLAB is wildly off compared to the correct bode plot. However, the Mathematica plot is quite close to the correct one. Here's the interesting thing. The transfer function whose bode plot is being taken in both the softwares is calculated to be the same. Have I made some mistake or is it numerical errors contributing to the problem? Here is relevant part of my code:

A=[0,0,0,0,1,0,0,0;
    0,0,0,0,0,1,0,0;
    0,0,0,0,0,0,1,0;
    0,0,0,0,0,0,0,1;
    -297.3,163.5,0,0,0,0,0,0;
    162.9,-267.2,104.2,0,0,0,0,0;
    0,57.8,-74.2,16.4,0,0,0,0;
    0,0,16.4,-16.4,0,0,0,0];
B=[0,0,0,0,0;
    0,0,0,0,0;
    0,0,0,0,0;
    0,0,0,0,0;
    131.4,0.046,0,0,0;
    0,0,0.045,0,0;
    0,0,0,0.025,0;
    0,0,0,0,0.25];
S=[1,0,0,0,0,0,0,0;
    0,0,0,1,0,0,0,0];
D=zeros(2,5);
sys=ss(A,B,S,D)
systf=tf(sys);
s1a1=systf(1,2);
bode(s1a1);

[MATLAB plot which is way off the correct one][1][Mathematica Plot which is quite close to the correct one]2

EDIT: The transfer function being plotted here is: $$\frac{0.046s^6-5.522\times 10^{-17}s^5+16.46 s^4 - 1.748\times 10^{-14} s^3 + 880.1 s^2 - 7.808\times 10^{-14} s + 7108}{s^8 - 4.373\times 10^{-15} s^7 + 655.1 s^6 - 2.258\times 10^{-12} s^5 + 9.887\times 10^{4} s^4 - 2.144\times 10^{-10} s^3 + 3.43\times 10^{6} s^2 - 4.353\times 10^{-9} s + 2.069\times 10^{7}}$$

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I suspect that the reason for the difference is the number of samples used in the plot, since the extreme points are likely going to infinity (or very large numbers) such that if less points are used, those large values are simply missed in the computation. If the OP uses the same number of points in each, then the same result should be achieved- and for all the points shown if they were super-imposed on each plot it would be clear that for the points given, the results are the same.

My first suspicion before the OP added the plots was that it was due to the units of the axis. That is not the case but leaving this part of my answer below as it is another common reason for Bode plot differences:

The units of the frequency axis should be confirmed as they could be in units of:

Continuous domain units of frequency:

Hertz, or cycles/sec

radians/sec

Discrete domain units of (normalized) frequency:

cycles per sample (unique domain is $[-1,1)$)

radians/sample (unique domain is $[-\pi, \pi)$)

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  • $\begingroup$ Ah sorry. I forgot to add the plots. I am editing my question to add them and hopefully they will help. The frequency units in both are different and so thats not a concern. My concern is the magnitude values being very wrong. $\endgroup$
    – ModCon
    Jun 20, 2020 at 12:26
  • 1
    $\begingroup$ @MadCon I see! and know why that is occuring, will update my answer $\endgroup$
    – Dan Boschen
    Jun 20, 2020 at 12:34
  • $\begingroup$ Thank you so much! Taking more number of points in the interval of interest did the trick. $\endgroup$
    – ModCon
    Jun 20, 2020 at 13:01
  • $\begingroup$ I have observed that just like less points is bad, too many points also gives wrong plot. It seems that somewhere in the middle gives the best 'accurate' plot. Do you know of such a thing? If yes, could you please tell why does this happen? $\endgroup$
    – ModCon
    Jun 20, 2020 at 16:08
  • 1
    $\begingroup$ Yes that is correct $\endgroup$
    – Dan Boschen
    Jun 21, 2020 at 12:16

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